# Relating the characteristic of the ring R to the characteristic of R[x]

Suppose $R$ is a ring and $R[x]$ is the ring of polynomials in the indeterminate $x$ with coefficients from $R$. The characteristic of a ring is the smallest positive integer $n$ such that $n \cdot r =0$ for all $r$ in $R$, or $0$ if no such $n$ exists.

I'm interested in the truth of statement "the characteristic of $R$ is equal to the characteristic of $R[x]$."

If $R$ has unity then I believe the statement is true. Likewise, if $R$ has characteristic $0$ I believe the statement is true. The last case is then rings with no unity and characteristic $n>0$. However, no examples of this kind come to mind.

My questions are then:

1. Is the statement true?

2. If so, is there an easier way to show it besides a case breakdown?

3. Are there rings with no unity and positive characteristic?

Thank you.

• Isn't this just a direct proof? If ever element of $nr=0$ for all $r\in R$, then for every element $p(x)\in R[x]$, $np(x)=0$. Since $R\in R[x]$, you'd have the obvious that the characteristic of $R[x]$ cannot be less than the characteristic of $R$... – Thomas Andrews Aug 20 '14 at 18:35
• Hmm, it seems to be less clear why $nx$ should be zero if $n$ isn't furnished as a ring element. – rschwieb Aug 20 '14 at 18:37
• $x\not\in R[x]$ @rschwieb unless $R$ has a unit element. – Thomas Andrews Aug 20 '14 at 18:38
• I suppose it depends on how you define $R[x]$ when $R$ does not have an identity element. Never had to think of that. @rschwieb – Thomas Andrews Aug 20 '14 at 18:40
• @ThomasAndrews It depends on how you're forming polynomials, apparently. For some people, it's the free ring generated by $x$ and $R$, but apparently your are thinking of formal monomials with coefficients from $R$. If it's the latter the problem is easy, and if it's the former I guess powers of $x$ don't have a predetermined additive order. – rschwieb Aug 20 '14 at 18:40

1) If you're defining $R[x]$ as the set of finite sums $\{\sum r_ix^i\mid r_i\in R, i\in \Bbb Z^+\}$ (as we imagine you are) then it is very obvious that if $nR=0$, then $np(x)=\sum nr_ix^i=\sum 0x^i=0$. Conversely if $nR[x]=0$, it holds for the subset $R$. So taking $n$ to be minimal, 1) is true.
3) $2\Bbb Z/4\Bbb Z$ is such a rng.